Question

If ${\tan }^{-1}2+{\tan }^{-1}3+\theta =\pi$, find the value of $\theta$.

Answer 1

$Let\theta =\frac{\pi }{4}$

${\tan }^{-1}1+{\tan }^{-1}2+{\tan }^{-1}3={\tan }^{-1}1+({\tan }^{-1}2+{\tan }^{-1}3)$

$=\frac{\pi }{4}+\pi +{\tan }^{-1}\left(\frac{2+3}{1-2\times 3}\right)$

$=\frac{\pi }{4}+\pi -\frac{\pi }{4}$

$=\pi$